Exercise 6.7.1 of Tao’s Analysis I In Exercise 6.7.1 of Analysis I, Tao asks (among several things) to prove that for a real number $x>0$ and real numbers $\alpha$ and $\beta$, $(x^\alpha)^\beta=x^{\alpha\beta}$.
I’m stuck because following his Definition 6.7.2, I can write (I’ve proved this) $(x^\alpha)^\beta=\lim\limits_{n’\to\infty}\lim\limits_{n\to\infty} x^{q_n r_{n’}}$ where $(q_n)_{n=1}^\infty$ and $(r_n)_{n=1}^\infty$ are rational sequences converging to $\alpha$ and $\beta$ respectively.
But Tao doesn’t mention anywhere before this section how to handle double limits, and I have no clue how to prove that this is equal to $\lim\limits_{n\to\infty} x^{q_n r_n}$ which is equal to $x^{\alpha\beta}$.
Maybe I’m on the wrong track. Any help would be great.
 A: You need Lemma 6.7.1 (p152) "Continuity of exponentiation" when he proves that the result of real exponentiation is independent of the sequence (of rationals) converging to $a$ (only the limit counts).\ 
First, remark that, if $\alpha$ or $\beta$ is zero, then $(x^\alpha)^\beta=x^{\alpha\beta}$ holds. 
Then, we first suppose $\alpha>0$ (resp. $\beta>0$) and replace $q_n$ (resp. $r_m$) by any increasing sequence of strictly positive rationals $\hat{q}_n$ (resp. $\hat{r}_m$) converging to $\alpha$ (resp. $\beta$). 
Due to the fact that the sequences are increasing, we have, for all $n,m$
$$
x^{\hat{q}_{n}\cdot \hat{r}_{m}}=(x^{\hat{q}_{n}})^{\hat{r}_{m}}\leq (x^{\alpha})^{\beta}\ ;\ 
(x^{\hat{q}_{n}})^{\hat{r}_{m}}=x^{\hat{q}_{n}\cdot \hat{r}_{m}}\leq x^{\alpha\cdot \beta}
$$ 
in the first one you take the limit $n\to\infty$ and get 
$x^{\alpha\cdot \hat{r}_{m}}\leq (x^{\alpha})^{\beta}$ then, with $m\to\infty$, you get 
$x^{\alpha\cdot \beta}\leq (x^{\alpha})^{\beta}$. Using the second one, taking the limits in the same order ($n$ then $m$) you get successively $(x^{\alpha})^{\hat{r}_{m}}\leq x^{\alpha\cdot \beta}$ and then 
$(x^{\alpha})^{\beta}\leq x^{\alpha\cdot \beta}$. 
You finish the cases, using $x^{-a}=\dfrac{1}{x^{a}}$.
Hope it helps !
